A particle moves along the $x$ -axis. The function $x(t)$ gives the particle's position at any time $t\geq 0$ : $x(t)=t^3-3t^2-9t+1$ What is the particle's velocity $v(t)$ at $t=2$ ? $v(2)=$
Answer: We have a function for the particle's position, and we need to find the particle's velocity. Since velocity is the rate of change of position, we need to find the derivative of $x(t)$. In other words, if $v(t)$ gives the particle's velocity at any time $t\geq 0$, then $v(t)=x'(t)$. Let's differentiate $x(t)$ to find $v(t)$ : $\begin{aligned} v(t)&=x'(t) \\\\ &=\dfrac{d}{dt}[t^3-3t^2-9t+1] \\\\ &=3t^2-6t-9 \end{aligned}$ To find the particle's velocity at $t=2$, we need to evaluate $v(2)$. $\begin{aligned} v({2})&=3({2})^2-6({2})-9 \\\\ &=-9 \end{aligned}$ Since the velocity is negative, we know the particle is moving to the left. The particle's velocity at $t=2$ is $-9$. At $t=2$, the particle is moving to the left.